1
$\begingroup$

I've come across this problem and I've no idea how to handle it: Show that the set of all group isomorphisms from $f:(Q,+)\rightarrow (Q,+)$ is isomorphic with the $(Q^*,\cdot)$ group. Anyone here could lend a hand, maybe an arm?

PS: This question has been market as a possible duplicate of the following: Automorphism group of $\mathbb{Q}$ considered as a group under addition. The person who asked the linked question was wondering if he could prove that $Aut(Q,+)$ is isomorphic with the $(Q^*,+)$ group, whether my question was on $Aut(Q,+)$ and $(Q*,\cdot)$. I hope this is enough of an explanation. Notice me if not.

$\endgroup$
2
$\begingroup$

Hint: Show that any group isomorphism $\varphi: (\mathbb{Q},+) \to (\mathbb{Q},+)$ is of the form $\varphi(x) = ax$ for $a\neq 0$. Clearly, then, composition of such maps is the same thing as multiplication.

To show this, let $a=\varphi(1)$, then show that $\varphi(n) = an$ for $n\in\mathbb{Z}$, then end with extending this to $\mathbb{Q}$.

$\endgroup$
1
$\begingroup$

First prove that every isomorphism $(\Bbb Q,+) \to (\Bbb Q,+) $ is of the form $\psi_x(y) = xy$ (where $x\in \Bbb Q^*$) and that every function of this form is an isomorphism

The rest follow immediatly

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.